I've tried to design a voltage controlled oscillator by adding a varactor to a Colpitts oscillator. enter image description here

However simulation shows no change in the oscillation frequency as V3 changes (0.04v to 10v). I don't know what is the problem with the design.

Here is the FFT of the output voltage: enter image description here

EDIT: Increasing C3 to 1nF (for 10 12 and 15 volts) enter image description here

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    \$\begingroup\$ try increasing C3 to 1nF or so. \$\endgroup\$ Commented Jan 23, 2016 at 9:57
  • \$\begingroup\$ @Jasen Increasing C3 changed frequency but the output is distorted. \$\endgroup\$
    – SMA.D
    Commented Jan 23, 2016 at 10:12
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    \$\begingroup\$ increase R4 to 1M, that will possibly help a bit. and keep V3 high enough that D1 is reverse biased during the full oscillation cycle \$\endgroup\$ Commented Jan 23, 2016 at 10:23
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    \$\begingroup\$ you hve the Varicap in the collecter circuit where it sees a higher signal voltage, it will probably work better in the emitter circuit where the voltage it sees will effect its capacitance less. \$\endgroup\$ Commented Jan 23, 2016 at 10:55

1 Answer 1


Trace out the parallel resonant circuit, L1 in parallel with a capacitance.

I see L1 -> (C1 and C2 in series to ground) in parallel with (C3 and D1 in series to ground) -> ... oops, the capacitance chain stops there and never gets back to L1.

First fix this by decoupling V2 to ground.

Secondly note that C3 limits the capacitance variability from D1 to at most 10% of the total, because C3=1pf while C1=10pf. So at this stage, increase C3 significantly - to something larger than C1 - perhaps 100pf.

When the circuit is working correctly but has excessive tuning range, then you can reduce C3 appropriately.

(R2 and R4 decrease the Q of the tuned circuit, increasing them or eliminating R2 altogether may help)


I don't understand this : "the capacitance chain stops there and never gets back to L1"

That's why I said it. (Not to be obtuse, but to focus attention here)

Look at the other end of L1. There is no low-impedance AC path to ground - at least on the schematic. And hence, no well-defined capacitance in parallel with L1, hence no tuned circuit. You can't rely on a battery or PSU to provide a low-impedance AC path.

If there is a decoupling cap on the prototype, good - but that's an error in the schematic. If there isn't, that's an error in both the schematic and circuit. Add such a capacitor from the L1 pin itself straight to the ground point where both C2 and D1 connect.

If C2 and D1 connect to ground some distance apart, that's another error on the circuit. For such low capacitances they need to be arranged with the smallest possible loop area around all capacitors and L1.

  • \$\begingroup\$ Why PSU can't be relied as a low-impedance AC path? \$\endgroup\$
    – SMA.D
    Commented Jan 23, 2016 at 15:15
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    \$\begingroup\$ How long are the connections to it, for a start? Those connections have inductance, probably more inductance than L1, for a start. \$\endgroup\$
    – user16324
    Commented Jan 23, 2016 at 15:35
  • \$\begingroup\$ I think You mean if the connections are far from the battery (producing non-negligible capacitance and inductance) a decoupling cap is required. Am I true? \$\endgroup\$
    – SMA.D
    Commented Jan 23, 2016 at 15:40
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    \$\begingroup\$ That's a start. But also, do you trust that a battery has low output impedance at RF? And if so, why? Either way, I would suggest a decoupling cap is a very good idea. \$\endgroup\$
    – user16324
    Commented Jan 23, 2016 at 15:51

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